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Get dimensions of an image on upload can this be done

Posted on 2007-03-17
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Last Modified: 2008-03-25
Is it possible without the use of imagemagick or GD or such like with php alone to get the dimensions of an image that is being uploaded?

I mean, lets say some user wants to upload an image 1000w x 600h pixels and we dont want an image that large on our server could this upload code we are using be modified to terminate the upload and give a warning dialog  stating the max dimensions we will allow?

Here is our code

<?

$i=0;




    if (is_dir ("gallery/".$userdata['username']."")) {
       
    }
      else
      {
         $mypath="gallery/".$userdata['username']."";
           mkdir($mypath,0777);
      }



if ($_FILES["file"]["size"] < 20000000)
      {
              if ($_FILES["file"]["error"] > 0)
                     {
                      echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
                }
             else
                {
                        echo "Uploaded: " . $_FILES["file"]["name"]. $i . "<br>";
                        while (file_exists("gallery/".$userdata['username']."/" . $_FILES["file"]["name"]. $i))
                            {
                                    $i++;
                                  }

                        move_uploaded_file($_FILES["file"]["tmp_name"], "gallery/".$userdata['username']."/" . $_FILES["file"]["name"] . $i);
                        echo "Stored in: " . "".$userdata['username']."/" . $_FILES["file"]["name"] . $i;
                        echo '<br><br>';
                }
              }
 
  $url =  "gallery/".$userdata['username']."/" . $_FILES["file"]["name"]. $i;
 
 
$uploadimage = 'INSERT INTO gallery (`user_name`, `user_id`, `image_url`, `image_desc`, `image_name`, `image_website`) VALUES (\''.$_POST['username'].'\',\''.$userdata['user_id'].'\', \''.$url.'\', \''.$_POST['desc'].'\', \''.$_POST['name'].'\', \''.$_POST['web'].'\');';
 mysql_query($uploadimage);

0
Comment
Question by:vbMarkO
5 Comments
 
LVL 15

Accepted Solution

by:
babuno5 earned 500 total points
Comment Utility
check this function out
http://in2.php.net/getimagesize
0
 
LVL 14

Expert Comment

by:raja_ind82
Comment Utility
------------------------------------------------------------------------
<?php
//$a_imagesize = getimagesize($_FILES['uploaded_filename']['tmp']);
$a_imagesize = getimagesize("Sunset.jpg");
print_r($a_imagesize);
?>
-------------------------------------------------------------------------------------
[0] = width
[1] = height
[2] = type
[3] = "width=x heigh=y" for the <img> tag.

Returns an array with 4 elements.

Index 0 contains the width of the image in pixels.
Index 1 contains the height.
Index 2 is a flag indicating the type of the image:
 1 = GIF,
 2 = JPG,
 3 = PNG,
 4 = SWF,
 5 = PSD,
 6 = BMP,
 7 = TIFF(intel byte order),
 8 = TIFF(motorola byte order),
 9 = JPC,
 10 = JP2,
 11 = JPX,
 12 = JB2,
 13 = SWC,
 14 = IFF,
 15 = WBMP,
 16 = XBM.
 These values correspond to the IMAGETYPE constants that were added in PHP 4.3.0.
Index 3 is a text string with the correct height="yyy" width="xxx" string that can be used directly in an IMG tag.

NOTE: Just because getimagesize can return types in a lot of images doesn't mean it can process them. TIFF files is a good example!

i hope this will help you.

Regards,
M.Raja
0
 
LVL 1

Expert Comment

by:gobinathm
Comment Utility
list($width, $height, $type, $attr) = getimagesize($file)

$file is nothing but the name of the file you have uploaded. By this method you can get it in seperate variables instead of getting all the values in an array.

Now you will have you width in the variable $width & height in the variable $height.
0
 
LVL 1

Expert Comment

by:Computer101
Comment Utility
Forced accept.

Computer101
EE Admin
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